3.1.29 \(\int (a+b \tanh ^{-1}(c x))^3 \, dx\) [29]

Optimal. Leaf size=108 \[ \frac {\left (a+b \tanh ^{-1}(c x)\right )^3}{c}+x \left (a+b \tanh ^{-1}(c x)\right )^3-\frac {3 b \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1-c x}\right )}{c}-\frac {3 b^2 \left (a+b \tanh ^{-1}(c x)\right ) \text {PolyLog}\left (2,1-\frac {2}{1-c x}\right )}{c}+\frac {3 b^3 \text {PolyLog}\left (3,1-\frac {2}{1-c x}\right )}{2 c} \]

[Out]

(a+b*arctanh(c*x))^3/c+x*(a+b*arctanh(c*x))^3-3*b*(a+b*arctanh(c*x))^2*ln(2/(-c*x+1))/c-3*b^2*(a+b*arctanh(c*x
))*polylog(2,1-2/(-c*x+1))/c+3/2*b^3*polylog(3,1-2/(-c*x+1))/c

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Rubi [A]
time = 0.16, antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {6021, 6131, 6055, 6095, 6205, 6745} \begin {gather*} -\frac {3 b^2 \text {Li}_2\left (1-\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c}+x \left (a+b \tanh ^{-1}(c x)\right )^3+\frac {\left (a+b \tanh ^{-1}(c x)\right )^3}{c}-\frac {3 b \log \left (\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{c}+\frac {3 b^3 \text {Li}_3\left (1-\frac {2}{1-c x}\right )}{2 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x])^3,x]

[Out]

(a + b*ArcTanh[c*x])^3/c + x*(a + b*ArcTanh[c*x])^3 - (3*b*(a + b*ArcTanh[c*x])^2*Log[2/(1 - c*x)])/c - (3*b^2
*(a + b*ArcTanh[c*x])*PolyLog[2, 1 - 2/(1 - c*x)])/c + (3*b^3*PolyLog[3, 1 - 2/(1 - c*x)])/(2*c)

Rule 6021

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTanh[c*x^n])^p, x] - Dist[b
*c*n*p, Int[x^n*((a + b*ArcTanh[c*x^n])^(p - 1)/(1 - c^2*x^(2*n))), x], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[p
, 0] && (EqQ[n, 1] || EqQ[p, 1])

Rule 6055

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-(a + b*ArcTanh[c*x])^p)
*(Log[2/(1 + e*(x/d))]/e), x] + Dist[b*c*(p/e), Int[(a + b*ArcTanh[c*x])^(p - 1)*(Log[2/(1 + e*(x/d))]/(1 - c^
2*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 - e^2, 0]

Rule 6095

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTanh[c*x])^(p
+ 1)/(b*c*d*(p + 1)), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[c^2*d + e, 0] && NeQ[p, -1]

Rule 6131

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTanh[c
*x])^(p + 1)/(b*e*(p + 1)), x] + Dist[1/(c*d), Int[(a + b*ArcTanh[c*x])^p/(1 - c*x), x], x] /; FreeQ[{a, b, c,
d, e}, x] && EqQ[c^2*d + e, 0] && IGtQ[p, 0]

Rule 6205

Int[(Log[u_]*((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(-(a + b*ArcT
anh[c*x])^p)*(PolyLog[2, 1 - u]/(2*c*d)), x] + Dist[b*(p/2), Int[(a + b*ArcTanh[c*x])^(p - 1)*(PolyLog[2, 1 -
u]/(d + e*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d + e, 0] && EqQ[(1 - u)^2 - (1
- 2/(1 - c*x))^2, 0]

Rule 6745

Int[(u_)*PolyLog[n_, v_], x_Symbol] :> With[{w = DerivativeDivides[v, u*v, x]}, Simp[w*PolyLog[n + 1, v], x] /
;  !FalseQ[w]] /; FreeQ[n, x]

Rubi steps

\begin {align*} \int \left (a+b \tanh ^{-1}(c x)\right )^3 \, dx &=x \left (a+b \tanh ^{-1}(c x)\right )^3-(3 b c) \int \frac {x \left (a+b \tanh ^{-1}(c x)\right )^2}{1-c^2 x^2} \, dx\\ &=\frac {\left (a+b \tanh ^{-1}(c x)\right )^3}{c}+x \left (a+b \tanh ^{-1}(c x)\right )^3-(3 b) \int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{1-c x} \, dx\\ &=\frac {\left (a+b \tanh ^{-1}(c x)\right )^3}{c}+x \left (a+b \tanh ^{-1}(c x)\right )^3-\frac {3 b \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1-c x}\right )}{c}+\left (6 b^2\right ) \int \frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx\\ &=\frac {\left (a+b \tanh ^{-1}(c x)\right )^3}{c}+x \left (a+b \tanh ^{-1}(c x)\right )^3-\frac {3 b \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1-c x}\right )}{c}-\frac {3 b^2 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{c}+\left (3 b^3\right ) \int \frac {\text {Li}_2\left (1-\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx\\ &=\frac {\left (a+b \tanh ^{-1}(c x)\right )^3}{c}+x \left (a+b \tanh ^{-1}(c x)\right )^3-\frac {3 b \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1-c x}\right )}{c}-\frac {3 b^2 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{c}+\frac {3 b^3 \text {Li}_3\left (1-\frac {2}{1-c x}\right )}{2 c}\\ \end {align*}

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Mathematica [A]
time = 0.31, size = 161, normalized size = 1.49 \begin {gather*} \frac {2 a^3 c x+6 a^2 b c x \tanh ^{-1}(c x)+3 a^2 b \log \left (1-c^2 x^2\right )+6 a b^2 \left (\tanh ^{-1}(c x) \left ((-1+c x) \tanh ^{-1}(c x)-2 \log \left (1+e^{-2 \tanh ^{-1}(c x)}\right )\right )+\text {PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c x)}\right )\right )+b^3 \left (2 \tanh ^{-1}(c x)^2 \left ((-1+c x) \tanh ^{-1}(c x)-3 \log \left (1+e^{-2 \tanh ^{-1}(c x)}\right )\right )+6 \tanh ^{-1}(c x) \text {PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c x)}\right )+3 \text {PolyLog}\left (3,-e^{-2 \tanh ^{-1}(c x)}\right )\right )}{2 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTanh[c*x])^3,x]

[Out]

(2*a^3*c*x + 6*a^2*b*c*x*ArcTanh[c*x] + 3*a^2*b*Log[1 - c^2*x^2] + 6*a*b^2*(ArcTanh[c*x]*((-1 + c*x)*ArcTanh[c
*x] - 2*Log[1 + E^(-2*ArcTanh[c*x])]) + PolyLog[2, -E^(-2*ArcTanh[c*x])]) + b^3*(2*ArcTanh[c*x]^2*((-1 + c*x)*
ArcTanh[c*x] - 3*Log[1 + E^(-2*ArcTanh[c*x])]) + 6*ArcTanh[c*x]*PolyLog[2, -E^(-2*ArcTanh[c*x])] + 3*PolyLog[3
, -E^(-2*ArcTanh[c*x])]))/(2*c)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(244\) vs. \(2(106)=212\).
time = 0.24, size = 245, normalized size = 2.27

method	result	size
derivativedivides	\(\frac {a^{3} c x +b^{3} c x \arctanh \left (c x \right )^{3}+b^{3} \arctanh \left (c x \right )^{3}-3 b^{3} \arctanh \left (c x \right )^{2} \ln \left (1+\frac {\left (c x +1\right )^{2}}{-c^{2} x^{2}+1}\right )-3 b^{3} \arctanh \left (c x \right ) \polylog \left (2, -\frac {\left (c x +1\right )^{2}}{-c^{2} x^{2}+1}\right )+\frac {3 b^{3} \polylog \left (3, -\frac {\left (c x +1\right )^{2}}{-c^{2} x^{2}+1}\right )}{2}+3 a \,b^{2} c x \arctanh \left (c x \right )^{2}+3 a \,b^{2} \arctanh \left (c x \right )^{2}-6 \arctanh \left (c x \right ) \ln \left (1+\frac {\left (c x +1\right )^{2}}{-c^{2} x^{2}+1}\right ) a \,b^{2}-3 \polylog \left (2, -\frac {\left (c x +1\right )^{2}}{-c^{2} x^{2}+1}\right ) a \,b^{2}+3 a^{2} b c x \arctanh \left (c x \right )+\frac {3 a^{2} b \ln \left (-c^{2} x^{2}+1\right )}{2}}{c}\)	\(245\)
default	\(\frac {a^{3} c x +b^{3} c x \arctanh \left (c x \right )^{3}+b^{3} \arctanh \left (c x \right )^{3}-3 b^{3} \arctanh \left (c x \right )^{2} \ln \left (1+\frac {\left (c x +1\right )^{2}}{-c^{2} x^{2}+1}\right )-3 b^{3} \arctanh \left (c x \right ) \polylog \left (2, -\frac {\left (c x +1\right )^{2}}{-c^{2} x^{2}+1}\right )+\frac {3 b^{3} \polylog \left (3, -\frac {\left (c x +1\right )^{2}}{-c^{2} x^{2}+1}\right )}{2}+3 a \,b^{2} c x \arctanh \left (c x \right )^{2}+3 a \,b^{2} \arctanh \left (c x \right )^{2}-6 \arctanh \left (c x \right ) \ln \left (1+\frac {\left (c x +1\right )^{2}}{-c^{2} x^{2}+1}\right ) a \,b^{2}-3 \polylog \left (2, -\frac {\left (c x +1\right )^{2}}{-c^{2} x^{2}+1}\right ) a \,b^{2}+3 a^{2} b c x \arctanh \left (c x \right )+\frac {3 a^{2} b \ln \left (-c^{2} x^{2}+1\right )}{2}}{c}\)	\(245\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c*x))^3,x,method=_RETURNVERBOSE)

[Out]

1/c*(a^3*c*x+b^3*c*x*arctanh(c*x)^3+b^3*arctanh(c*x)^3-3*b^3*arctanh(c*x)^2*ln(1+(c*x+1)^2/(-c^2*x^2+1))-3*b^3
*arctanh(c*x)*polylog(2,-(c*x+1)^2/(-c^2*x^2+1))+3/2*b^3*polylog(3,-(c*x+1)^2/(-c^2*x^2+1))+3*a*b^2*c*x*arctan
h(c*x)^2+3*a*b^2*arctanh(c*x)^2-6*arctanh(c*x)*ln(1+(c*x+1)^2/(-c^2*x^2+1))*a*b^2-3*polylog(2,-(c*x+1)^2/(-c^2
*x^2+1))*a*b^2+3*a^2*b*c*x*arctanh(c*x)+3/2*a^2*b*ln(-c^2*x^2+1))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))^3,x, algorithm="maxima")

[Out]

a^3*x + 3/2*(2*c*x*arctanh(c*x) + log(-c^2*x^2 + 1))*a^2*b/c - 1/8*((b^3*c*x - b^3)*log(-c*x + 1)^3 - 3*(2*a*b
^2*c*x + (b^3*c*x + b^3)*log(c*x + 1))*log(-c*x + 1)^2)/c - integrate(-1/8*((b^3*c*x - b^3)*log(c*x + 1)^3 + 6
*(a*b^2*c*x - a*b^2)*log(c*x + 1)^2 - 3*(4*a*b^2*c*x + (b^3*c*x - b^3)*log(c*x + 1)^2 - 2*(2*a*b^2 - b^3 - (2*
a*b^2*c + b^3*c)*x)*log(c*x + 1))*log(-c*x + 1))/(c*x - 1), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))^3,x, algorithm="fricas")

[Out]

integral(b^3*arctanh(c*x)^3 + 3*a*b^2*arctanh(c*x)^2 + 3*a^2*b*arctanh(c*x) + a^3, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \operatorname {atanh}{\left (c x \right )}\right )^{3}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(c*x))**3,x)

[Out]

Integral((a + b*atanh(c*x))**3, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))^3,x, algorithm="giac")

[Out]

integrate((b*arctanh(c*x) + a)^3, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )}^3 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x))^3,x)

[Out]

int((a + b*atanh(c*x))^3, x)

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